Imagine you are flipping a two-faced coin (heads and tails). Lets say you earn 1 point for every win and lose 1 point for every loss. You have the option of flipping the coin without winning or losing any points (neutral flips). How would you maximize the rate you win points?
Personally, I would flip heads until I get two losses in a row, then do neutral flips until the series of tails ends. Then continue flipping heads and repeat the process. This eliminates the possibility of me getting a large losing streak and guarantees me getting everything I can get out of a winning streak.
it doesn't matter what you pick, the past doesn't influence the future flips and there is no preference in a fair coin toss.
nb4 >gamblers fallacy
Easton Fisher
of course your past doesn't influence future flips. but you cant deny that due to the nature of randomness, there will be long streaks of wins and long streaks of losses. Using this system, i was able to win over 2500 points on a similar game i played.
If you are making bets on each flip separately, the chance is always 50/50
You can hook up the sequence into one piece by constantly doubling the bet, but this needs insanely deep pockets - also, the House knows this also and that's why they *always* limit the size of a bet
Xavier Cooper
is correct. Past flips have no effect on future outcomes. Whatever happened before, it's 50/50 what happens next. Common psychology experiment. Half a group is asked to flip a coin 100 times and write HHTHTTTTH on a blackboard. The other half is told to _imagine_ flipping a coin and writing down the sequence. The instructor re-enters the room and easily picks out which sequences are the result of a real coin.
Simple computer program using a random-generator. There are runs of 5, 6, and 7 consecutive tails. The "imaginary" sequences never include such "streaks" because people _think_ a tail is more likely to be followed by a head, and vice versa.
Dylan Adams
Your logic is similar to the fallacy that a random walk (either left of right) will invariably bring you back to your starting point. Not true, As the number of steps you take increases you are more and more likely to be found _further_ from the start. But the probable displacement only grows as the square-root of the number of steps. So the percentage "error" -- abs(heads-tails)/(heads+tails) -- decreases.
If you're convinced you've got a winning system though, don't let us stop you. Try Day Trading.
Nicholas Green
i know it is always 50/50. I never said i was changing the probability. But like i said, due to the nature of randomness, there will be times where heads or tails will be flipped several times in a row. My system isnt trying to change any probability, it is only avoiding the bad "streaks". how can you say streaks are imaginary when the program very clearly shows both sides at times being flipped in a row several times?
Lucas Howard
50/50 either way 1/4 for two same in series
Adam Morales
Again, my system doesn't try to change any probability. It only avoids the series of undesired results that occur due to the nature of randomness.
Camden Lee
you don't know in advance what the next flip is, and no choice is preferred, so there is no strategy except dumb luck.
Elijah Murphy
Can't be done. If there was a way, the sequence wouldn't be random.
Bentley Allen
read my strategy. it has noting t do with knowing what flip is next. my system works BECAUSE it is random. it doesn't change probability and it doesn't try to predict the future.
Evan Hill
your "system" assumes that the past matters. Therefore it is the gambler's fallacy.
Andrew Johnson
I see what you're saying.
Start counting points on a heads. And stop counting after the second of a 2 heads streak or if it's tails. Resume counting on the next heads flip.
You eliminate tails streaks this way, while banking on heads streaks, so statistically you will be gaining points, very slowly.
Justin Scott
no because the probability of flipping a series of ten heads is the same as flipping a series with five heads and five tails
take a fucking stats class brainlets
Luke Miller
it doesnt assume that the past matters it only assumes that HH is more likely than HTHT
Brody Lopez
this would suggest you get to count the head you flip after you see that it's heads
every couple flip is a 50/50 chance regardless of the previous flip(s). therefore you are just as likely to get the same number of heads when counting every flip as you would pausing after 2 bad ones and resuming after the next hour
there is zero statistical gain or loss in that system. you simply exclude some flips. which is the same result as if you just flipped that many times less in total
Carson Harris
and that fact is pointless because the shit that already happened doesn't matter when it comes to what the result of the next flip is going to be. Wipe the slate clean, CLEAN, after every flip. Nothing that happened matters or influences the next flip. Trying to include the next flip in some probability about the next flip is the definition of the gamblers fallacy.
Aiden Myers
No, you dont. If you see it is heads, you start going for points. It doesnt include the first heads you see after doing neutral flips.
Brandon Bailey
The fact that it's 50/50 chance doesn't matter. You're eliminating every single possibility of a tail streak, while allowing any random HH streak to give you a point.
The only way you lose points with this system is HTHT a 4 number sequence. You gain points with HH a 2 number sequence. A 2 number sequence is more likely than a 4 number sequence in any random series. Just like HH is more likely than HHHH.
Yes, HH and TT have the same likelihood. HH and HTHT (it's counter in this system) do not.
Gabriel Parker
>why casinos stay in business the post
Eli White
Did an experiment.
Daniel Johnson
>this game is exactly like a casino lmao no, but nice shitposting
Dominic Rodriguez
There are "streaks" but you never know you're in one. After two tails you go neutral and stay neutral until you get a head. Can't you see you've implicitly assumed you've "used up tails" and a streak of heads will follow.
Here. I'll run my coin flipper a dozen times. THHTTHHHHHTHHHTHTTTHHHHTHTHTHTHHHTHTTTTTTTHTHHHHHHHTTTHHTTHTHHTHHHHHHHTTHTTHTHHTTTHTHHTTHTTHHTTTHHTT TTTHTTTHHHTTTHHTHTTHTTTHHTHTHTTTTHHTTHTTHTHTHTHTTHHHHTTHTHTHTTTHHHTHHTTHHTHTTTHHHHHHHTHHTHTHTHHHTTTT TTTTHTTHTTTTTHHTHHTTTHHHTTTTTHHTTTTTHTHHTTTTHTHTHTTHHHHTHHTHHHTTTTTTHHHHHHHTTHHTHHTTHHHHHHTTTHTTTHTT TTHHHHTTTTTHHTTHTHTTHTHTTTHHTHHTHHTTHTHHHHHHTTHTTTHHTHTTHHTTHTHHTTTHHHTHTTTHHTHHHHHHTHHTHHTHTTTHHHHH THHHHTTHTHTHTHTHHTTHTHTTHTHTTHTTTHTTHHTHHTHTTTTHTHHHHHTTTTTTHTTHTHHTTTHHHHTHHTHHTHHTHHHHHTTTTTTTHTTH HTHTHTTTHTHHHHTHHTTTHHTHTTTHHHHTTHTTTHHHHHTHTHHHHHHTHHTTHTTHTTTTHHTTHTHHHTHTTHHTTHHHTHHTHTHTTTTHTHHT THHTHHHHTTHTTTHHHHHHTHTTHHTTHHTHTTTTTTTTTHTHHHTTTHHTTHTTHHHHTHHTHTTHHHHHTHHHTHHHTTTHHTTHTHTHHHTHHHHH HTHTHTHHTTTHHHTHHTTHTHTTTHTTTHTHTTTTHHHHTTHHTTHHHHHTTHHHHTTHTHHHHHHTTHTHTTHTHHTTTHTTHTHTHHTHHTHHTTTH HTTTTHHTHHHHTHTTTHHHHHTTTTTTHHTHHHTTHHTTTTTTHTTHHTHHHHHHTTTHHHHHTHTHHTHTTTTHHHHHHTHHTTTTHHTHTHTTHTHH HHTHTTTHHHTTTTTHTTHTTTTTHHHHHHTTTHTHHHTHHHHHHTHTHHTTHHHHHTTHTTTHHTTHTHHTHHTTTTTTTHTTTTHHHTHTTHHTHTTH TTHHHHTTHHHHHTTHTTHHHTHHTHTTTHHHHTHTTTHHHTHTHHHTTHTHHHTTHHTHHTHHHTHTTTHHTHTHTTHTHTHTTHTHHTTHHTHHHTHH THHHHTHHTHHTTHTTHHTTHTTTTHTHHTTTTTTTTTHTTHTHTTTHHTTHHTTHTTTTHTTTTHTTTTHHTHTTTHTTHHTTHHHTTTTTTTTTHTHH
How does your system work?
Brody Ramirez
dont know how to read this, cant tell if you are implementing the strategy correctly
Andrew Rivera
Count the number of HH vs HTHT
see:
Levi Harris
>trips nice >"used up tails" who are you quoting? also what does "used up tails" mean? And it doesnt assume there is a streak of heads, it continues until there is two tails in a row. >How does your system work? Um, you just said it here: >After two tails you go neutral and stay neutral until you get a head.
Josiah Ward
Basically you've got a state-space consisting of four variables: the current number of flips (i), the second to previous flip (x), the previous flip (x'), and the current total (total). During each iteration you flip a coin and If x and x' were not both losses you add it to your current point total and set it as x'. You keep doing this until you reach the target number of flips (n).
Jonathan Reyes
>Your logic is similar to the fallacy that a random walk (either left of right) will invariably bring you back to your starting point. For a perfectly fair walk, this is actually true. [math] p_0(2n) = {2n \choose n}\frac{1}{2^{2n}} [/math] Is the probability that after [math] 2n [/math] steps your 1D random walk is at the origin. We ignore the odd-numbered steps because if the particle starts at the origin, [math] p_0(2n+1) = 0 [/math] If we sum this probability from n=1 to infinity, we will get the expected number of returns to the origin. We can use Stirling's approximation to get the probability into a nicer form [math] p_0(2n) \approx \frac{1}{\sqrt{\pi \cdot n}} [/math] which is bounded below by [math] \frac{1}{n} [/math] for [math] n > 3 [/math], whose sum diverges. Therefore the expected number of returns to the origin is infinite, which implies that the simple random walk will always return to the origin.
Of course, this argument relies on a body of theory (one that I don't have the characters, eloquence, confidence, or understanding to explain) to justify that "the expected number of returns to the origin is 'infinity' implies you will always return to the origin." If you want the real nitty gritty justification, stackexchange (where I got the majority of this proof) recommends Rick Durret's "Probability Theory and Examples"
Luis Perry
Not quoting anyone. Quotes indicate metaphor.
By "how does your system work" I meant "please apply it to this sample data and tell us what your results were." If YOU do it, you can't complain someone else might not be implementing the strategy correctly.
Ian Gray
adding a point whenever you weren't wrong for two flips is not the same as the system described.
Jose Gray
>By "how does your system work" I meant "please apply it to this sample data and tell us what your results were." Then next time, just say "apply it to this sample data and tell us what your results were" Also idk how to do this. I use this >complaining and wanting to be sure the evidence isnt bias/false are the same thing lmao okay
>I would flip heads until I get two losses in a row, then do neutral flips until the series of tails ends. Then continue flipping You are waiting for two tails and then not adding anything up to and including the flip where you get heads. From there on you add the appropriate amount until you get another two tails.
Samuel Gomez
Yes I know that. But after an infinite time, we're all dead. The fallacy I was trying to point out is that, having drifted away from 0, the next N steps are no more likely to move you towards 0 than they are to move you away. Every set of steps can be considered as a brand new random walk. It has no memory of your original starting point.
Yes, you will always return to zero sometime before you've taken an infinite number of steps. This is exactly analogous to the strategy of gambling (a coin flip will do) and doubling the stakes after each bet. In theory, you're eventually guaranteed a fortune. In practice, this only works if you have an unlimited bankroll. The odds are that you'll go bankrupt (be unable to cover your bets) before the casino does.
Ian Sullivan
>H(point gained, stop)TT HOLY FUCK YOU CAN SEE INTO THE FUTURE??????
Justin Parker
Interesting. What language is that? Very compact. Anyway, I assume you've taken my data and shown that after 1200 tosses you're only ahead 3.
Matthew Watson
no, that's the point of the system. to eliminate the inevitable tails streaks.
You only lose on THT.. but you win on HH a more likely combo. You also win on any length of HH over 3, whereas this can never happen with T.
the fact that it's random is why you can win by doing it this way. op was not lyin'
Julian Brown
>then not adding anything up to and including the flip heh?
Austin Gray
Just think about it logically.. can HHHH occur? Yes. Can TTTT occur? No.
That alone you should be enough for you to realize you will come out on top of any arbitrary length of flips.
Hunter Foster
You can't flip and then decide whether or not you want to add the result. Otherwise you could just flip and only count it if it's heads.
Hard to believe there are people as dumb as OP out there.
Liam Morales
>the goyim have learned our system, shut it down!!!
Justin Johnson
Also I don't have a compiler on this computer or I'd have done it with a script u cheeky bugger
Parker Peterson
Icing on the brainlet cake.
Brayden Foster
>TH(start)H(point gained, stop)TTH(start) No, this should actually be: (Start)T(loss)H(point)T(loss)T(loss, start)H(point) > that's the point of the system. to eliminate the inevitable tails streaks. yes >You only lose on THT.. but you win on HH a more likely combo. No, you lose on TT also. And no result is any more or less likely than the other. The likelyhood of the results has nothing to do with the system. It will always be 50/50, which is why the system works.
Brody Wright
>You can't flip and then decide whether or not you want to add the result You don't.. you just count every flip after an H. You don't count anything after a T. The fact that it's random is exactly why it works. Because streaks are possible, and we can control T streaks.
Adam Hill
see
William Richardson
also compare the result with randomly selecting the same proportion of flips from the same data set. Or you can have a system that makes you a winner 50% of the time - count an odd number of flips
Adam Walker
Wasn't being sarcastic. Trying to understand your code.
Levi Fisher
No it should be
(start, total = 0) T -> (total = -1) H -> (total = 0) H -> (total = 1) T -> (total = 0) T -> (total = -1) **ZOMG LOSING STREAK DETECTED WEEWOOWEEWOO NOT ADDING ANYMORE** H -> (total = -1 [neutral flip]) **WTF??? LOSING STREAK OVER I GUESS....** H -> (total = 0) etc. etc.
>you just count every flip after an H. You don't count anything after a T Well that's different than your original strategy now isn't it? And it still doesn't work since you're just as likely to discount an H as you are to discount a T.
If I ever become world dictator basic probability questions will be mandatory and people who get them wrong will be euthanized.
Joseph Cox
this is why red/black bets on roulette aren't 1:1
Charles Lopez
...
Christopher Jones
Wrong. You can go TT stop. You lose 2 point. Neutral flip heads. Flip TT stop. You lose 2 more points. You're down 4 points because neutral flipping the first heads reduces your streak by 1. There's no net here on average.
Parker Johnson
>maybe if i over-exaggerate they will believe me No, surprise surprise it just makes you look like an artist. Also, POTENTIAL losing streak. it isnt predicting the future. >etc. etc. actually, your cherry picked example isn't the only thing that happens when these numbers are randomly generated.
Aaron White
You stop after every point gained from an H and you don't start betting until you get H again. It's impossible to get TT because you would never bet after a T.
This is the optimized version of OP's.
Hudson Mitchell
Flip coin, if heads, decide next flip will be neutral. Swap back and forth every flip.
Jacob Ross
pic congrats on being the only person who actually answered the question initially proposed in this thread
Grayson Sanchez
That's not optimal. If you stop after every heads, why would you start after a neutral heads?
Think about it in pairs. 25% for 2, 25% for -2, 50% for 0. If you flip 100 time, you stop at +3.
Jackson Carter
One of the biggest baits of all time, well done OP
Ryan Clark
idk about optimal for points but it eliminates big swings by eliminating any T streak and it's easier to understand.
Dominic Bennett
idk what you are talking about :)
Carson Hughes
Let's reduce this down to its simplest form. 1. Before each flip you make a decision to bet or not to bet. Right? 2. Your decision is based on the outcome of the last few flips -- regardless of whether you bet or not on them. Right? 3. If you decide not to bet your expected gain is zero. Right? 4. If you decide to bet it's 50-50 of landing heads so your expected gain is zero. Right? 5. So, whether you bet or don't bet, your expected gain is zero. Right?
Please point out exactly which numbered statement you don't agree with. And why.
Jeremiah Price
3. if you dont bet, it WILL be zero, it isn't an expected gain. 4. there is no expected gain here. it will be either +1 or -1.
Ryder Clark
And since both outcomes of 4 are equally likely, your net gain is zero. Expected net gain is _defined_ as what you might win * probability of winning - what you might lose * (1-probability of winning.)
I bet you a fair die will come up snake eyes. If it does, you pay me $20. If it doesn't, I pay you $3. My expected gain is 20*(1/6) - 3*(5/6) = $0.83 I will be happy to play as long as you like.
An expected gain is not THIS or THAT. It's one number -- what you can expect (over the long run.)
They are all equally likely and they sum to 0, meaning your strategy doesn't work.
Aaron Ward
It doesn't eliminate any streak. Here's how it would go: T stop. Flip heads neutral. Start again. Flip tails, stop. Neutral flip again.
There's no guarantee that you won't just get a steak of tails.
Ryder Walker
Here is an easy solution OP.
Write a program which takes a stream of bits (only one at a time, no looking ahead) and implements your strategy. Post it here.
Then I will feed it random bits and post the results here.
Jeremiah Rogers
Your theory is wrong because you include bets after losing streaks.OP's model would pull out until the losing streak is over, thus gaining a slight edge by choosing only winning streaks.
Jose Carter
writing simple program shows that op is full of shit
An even SIMPLER analysis. H and T are just labels. Your system would work equally well if you interchanged _all_ the Hs and Ts in your description. Some Other Guy (SOG) decides to use it. He gains a point when the coin lands T, loses one when it's H.
You and SOG play against each other. A coin flip only counts if BOTH OF YOU AGREE IN ADVANCE THAT IT DOES!
On any given toss there are two possibilities. 1. One or both of you, based on past results, decide not to bet. No money changes hands. If there's _never_ a flip when you both agree, then neither one of you profits. So we have to assume that AT LEAST SOMETIMES, you both agree.
2. You both agree the flip counts and the coin is tossed. Either you or SOG hands over a dollar and both of you decide strategy for the next toss.
If there's really a "winning system" then BOTH of you are making money since you're both using it. But that's silly. Both of you can't be getting rich. We have reached a contradiction. Some assumption must be wrong. Either the coin really isn't fair or THERE IS NO WINNING SYSTEM!
Julian Flores
Your coinfilp situation (without having the option of flipping the coin) is a martingale. Applying a betting strategy to a martingale (i.e doing neutral flips in your case) makes it stay a martingale. You cannot maximize the rate of winning on a martingale.
James Gray
Thanks for the word. I alluded to this in but didn't know the term.
Kayden Lewis
Why don't you write a computer program and try out different strategies?
>computer random number generation is deterministic! Yes, that's kind of unavoidable, but you can do a really good job of minimizing it with the C++ standard template library random number generator
Lucas Martin
Throwing that out there: it's possible to cheat at coin tosses but it takes some training. If you know your basic magic sleight of hand you should be able to flip a palmed coin easily. I'm using a 100 won coin I learned to rcognise the side of by palm touch only (it's quite easy with this coin and it flips well too). The rest is just flawless execution. You're probably wary or betting money on anyone that would offer you a coin toss bet, but there are times it came in handy netting me the best room in a colocation and the best house in inheritanc.
Jack Moore
Can we change this from "coin flips" to "Roulette wheel with only red and black"? It's the same thing (we removed 0 and 00 and there are 50 red and 50 black), it's just way sexier that way.
Start posting your program code and the results
Cooper Nelson
I don't think OP means "how would you cheat", he means "what strategy would you use" with an assumption that a benevolent AI robot is doing the flipping for the amusement of his human pets.
Mason Murphy
We've already provided random HHTHTHTTTHHHTHT strings and explained it, simply, half-a-dozen ways. If he's not convinced by now, it's hopeless.
Benjamin Brown
Most successful troll atempt of all time. Well done.
Ayden Jenkins
why cant you post code on this gay board
Oliver Flores
Let X be a discrete random variable denoting the total number of points earned. X ~ Bin(n, 1/2)
Jose Martinez
Is this a troll or do you know nothing about probability OP?
flipString = ''.join([random.choice("HT") for x in xrange(N)])
profit = 0
canBet = 1 lossesInRow = 0
for flip in flipString: if flip == 'H': profit += canBet lossesInRow = 0 canBet = 1 else: profit -= canBet lossesInRow += 1 if lossesInRow == 2: canBet = 0
print "profit per flip:", profit / float(N)
William Bell
With alternating blacks ad reds or groups of blacks and reds, like semi-circles or quadrants?
Daniel Williams
in theory its 50/50 and you could have long streaks of just heads, but in reality this is unlikely and after around ~3 heads it's likely to be tails next
Owen Diaz
I heard in stats class that there are long strings of the same result in the distribution of coin flips so there might actually be something to this
Gavin Cruz
>and after around ~3 heads it's likely to be tails next -_-
Lincoln Hall
huh.
It actually looks like you might have inadvertently proven op right.
This could be big.
James Hughes
There are long strings. See NOT getting occasional "runs" is a dead giveaway that the data is faked. But the system doesn't work. NO system can work. I posted and , either of which should convince you. Past behavior of a coin tells you NOTHING about the next toss. If you still think this is a money-maker, I suggest talking to your stats teacher.
Carter Bailey
Right
Cooper Hernandez
Run 100 flips 10 times looking for patterns and frequency of repeated back to back heads Use frequency to determine betting/neutral behaviour during real 100 flip game.
this can range from half the base score of no neutrals to double the base score of no neutrals, which in 10K flip sets is still neglible to the tune of 60-200pts out of a maximum of 5,000. It ignores frequency of chance repetition and fails to bet on roughly half of all heads.
Michael Bailey
Another implementation. Hopefully more readable.
Caleb Peterson
>93 replies to one of the most basic topics in all of mathematics So /pol/ was right, you guys really are fucking idiots
Oliver Anderson
Use biased coins.
Luis Hill
Yes that's right you should head on back so you don't catch our stupid.
Noah Scott
There IS no pattern of back to back head!!!!!!!!!!!!!!! That's what "random" means.
You're right -- insofar as too much effort has been expended to convince some people (the "fucking idiots") who just don't get it.
Nicholas Sanders
there is more metal in the tails side so you should alwasy bet on tails
Connor Cox
If the tails side is heavier, you should always bet HEADS since that will be the upper surface when it lands.
